Question

On a coordinate plane, triangle A B C is shown. Point A is at (0, 0), point B is at (3, 4), and point C is at (3, 2). What is the area of triangle ABC?

Answer 1

3

Explanation:

Answer 2

The area of triangle for the given coordinates is 1.5
`√(4.6)`

Explanation:

Given coordinates of triangles as

A = (0,0)

B = (3,4)

C = (3,2)

So, The measure of length AB = a =
`\sqrt{(x_2-x_1)^(2)+(y_2-y_1)^(2)}`

Or, a =
`\sqrt{(3-0)^(2)+(4-0)^(2)}`

Or, a =
`√(9+16)`

Or, a =
`√(25)`

∴ a = 5 unit

Similarly

The measure of length BC = b =
`\sqrt{(x_2-x_1)^(2)+(y_2-y_1)^(2)}`

Or, b =
`\sqrt{(3-3)^(2)+(2-4)^(2)}`

Or, a =
`√(0+4)`

Or, b =
`√(4)`

∴ b = 2 unit

And

So, The measure of length CA = c =
`\sqrt{(x_2-x_1)^(2)+(y_2-y_1)^(2)}`

Or, c =
`\sqrt{(3-0)^(2)+(2-0)^(2)}`

Or, c =
`√(9+4)`

Or, c =
`√(13)`

∴ c =
`√(13)` unit

Now, area of Triangle written as , from Heron's formula

A =
`√(s* (s-a)* (s-b)* (s-c))`

and s =
`(a+b+c)/(2)`

I.e s =
`(5+2+√(13))/(2)`

Or. s =
`(7+√(13))/(2)`

So, A =
`\sqrt{(((7+√(13)))/(2))* ((((7+√(13)))/(2))-5)* ((7+√(13))/(2)-2)* ((7+√(13))/(2)-√(13))}`

Or, A =
`\sqrt{(((7+√(13)))/(2))* (((√(13)-3))/(2))* ((4+√(13))/(2))* ((7-√(13))/(2))}`

Or, A =
`(3)/(2)` ×
`\sqrt{1+√(13) }`

∴ Area of triangle = 1.5
`√(4.6)`

Hence The area of triangle for the given coordinates is 1.5
`√(4.6)` Answer